Net 48N Force Stretches .5m Spring to .78m

In summary, to set up a spring with a length of 0.5m and a spring constant of 133.3N/m, a 48 Newton force must be exerted to stretch the spring. This will result in a total length of 0.9m (0.5m original length + 0.4m extension). To find the spring constant, use the formula F/k = x, where F is the force and x is the length of extension. Using a 4 Newton force and 0.03m extension, the spring constant is calculated to be 133.3N/m. To find the total length at 48 Newton force, use the formula F/k = x, where F is the
  • #1
c4iscool
31
0
how would I set this up: a spring that is .5m in length elongates by .03m when a 4 Newton force is exerted to stretch the spring. what is the total length of the spring(in m) when a net 48 Newton force is used to stretch the spring?
 
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  • #2
F=kx where x is the length of extension, k is the spring constant. From here you can find k right? With the k you can find for the extended length at 48N and not forgetting the original length.
 
  • #3
k = F/L. that would make k=8?
 
  • #4
I'm not sure how you got 8 but it's not correct. To find k, F/x => 4/0.03 and so the spring constant is 133.3N/m. From here are you able to continue?
 
  • #5
o, i c. i was using .5m as x.
 
  • #6
now that I have the constant, it should look like this:
f=k*x
48=133.3*x and I solve for x?
 
  • #7
You canot use 5m to find the spring constant, because at 5m, it is the natural length of th spring, and there is no force acting on it to provide the 5m.

Yes, and this x you get is only the length of extension, and the question wants the total length so I guess it's pretty easy from here!:wink:
 
  • #8
thanks for your help
 

What is net 48N force?

Net 48N force refers to the overall force acting on an object in a specific direction. It is the sum of all the forces acting on the object, including any applied force and forces due to gravity or other external factors.

How does a force of 48N stretch a spring from .5m to .78m?

When a force of 48N is applied to a spring, it causes the spring to stretch or compress. The amount of stretch or compression is directly proportional to the force applied, according to Hooke's law. In this case, the spring is stretched from its original length of .5m to .78m due to the applied force.

What is the significance of the .5m and .78m measurements in this experiment?

The .5m and .78m measurements represent the initial and final lengths of the spring, respectively. These measurements are important in determining the amount of stretch or compression that occurs in the spring due to the applied force of 48N.

Can the spring be stretched to a length greater than .78m with a net force of 48N?

It is possible for the spring to be stretched to a length greater than .78m with a net force of 48N, depending on the spring's elasticity and the material it is made of. However, the amount of stretch may not be directly proportional to the applied force and could lead to permanent deformation of the spring.

What are some real-world applications of this experiment?

This experiment can be applied in various fields, such as engineering, physics, and material science. For example, it can help determine the strength and elasticity of different materials, as well as how they respond to external forces. This information can be used in designing and testing various structures and products, such as bridges, buildings, and springs in mechanical devices.

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